Transcript 20. Electric Charge, Force, & Field

29. Maxwell’s Equations & Electromagnetic Waves
1.
2.
3.
4.
5.
6.
7.
8.
The Four Laws of Electromagnetism
Ambiguity in Ampere’s Law
Maxwell’s Equations
Electromagnetic Waves
Properties of Electromagnetic Waves
The Electromagnetic Spectrum
Producing Electromagetic Waves
Energy & Momentum in Electromagetic Waves
How does a conversation travel between cell phones?
By EM waves.
29.1. The Four Laws of Electromagnetism
4 Laws of EM (incomplete)
Law
Mathematical
Statement

Gauss for E

Gauss for B
Faraday
Ampere
(Steady I only)

dA 
E
B
0
dA  0
dr  
E
q
d B
Physical
Meaning
How q produces E;
E lines begin & end on q’s.
No magnetic monopole;
B lines form loops.
Changing B gives emf.
dt

B
dr  0 I
Moving charges give B.
Note E-B asymmetry between the Faraday & Ampere laws.
29.2. Ambiguity in Ampere’s Law
B in a RC circuit.
Ampere’s law:

B
C
dr  0 I
I is current through any open surface S bounded by C.
Current flows through surfaces 1,2,& 4.
But not 3.
 Ampere’s law fails ( for non-steady current ).
Maxwell’s modification:

0
B
C
dE
dr  0 I  0  0
dE
dt
 D isplacem ent curre nt
dt
Changing E gives I , which in turn gives B.
Example 29.1. A Capacitor
A parallel plate capacitor with plate area A and spacing d is charging at the rate dV/dt.
Show that the displacement current is equal to the current in the wires feeding the capacitor.
Capacitor :
q C V
Displacement current :
I 
dq
dV
C
dt
dt
E  E A
Id  0

dE
dt
V
(Current feeding capacitor)
A
d

0 A dV
d
dt
C
dV
dt
 I
QED
29.3. Maxwell’s Equations
Law
Mathematical
Statement
Gauss for E

E
Gauss for B

B

Faraday
AmpereMaxwell

B
E
q
dA 
dA  0
dr  
Physical
Meaning
How q produces E;
E lines begin & end on q’s.
0
No magnetic monopole;
B lines form loops.
d B
Changing B gives emf.
dt
dr  0 I  0  0
dE
dt
Maxwell’s Eqs (1864).
Classical electromagnetism.
Moving charges &
changing E give B.
Maxwell’s Equations in Vacuum
Gauss for E

E
dA  0
Gauss for B

B
dA  0
Faraday

E
dr  
Ampere-Maxwell

d B
dt
B
dr  0  0
dE
dt
29.4. Electromagnetic Waves
Faraday’s law:
changing B gives E.
Ampere-Maxwell’s law:
changing E gives B.

Electromagnetic (EM) waves
Plane Electromagnetic Wave
EM wave in vacuum is transverse: E  B  k (direction of propagation).
kˆ
EB
Right-hand rule
For uniqueness, see Prob 46
Sinusoidal plane waves going in x-direction:
E  x , t   E y  x , t  ˆj
 E p sin  kx   t  ˆj
B  x , t   B z  x , t  kˆ
 B p sin  kx   t  kˆ
Gauss’s Laws
Plane wave :
E  x , t   E y  x , t  ˆj
B  x , t   B z  x , t  kˆ
 E p sin  kx   t  ˆj
 B p sin  kx   t  kˆ
Both E & B field lines are straightlines,
so their flux over any closed surfaces vanish identically.
Hence the Gauss’s laws are satisfied.
Faraday’s Law

E
dr  
d B

dt
   E  dA
 
 B
 t
 dA
For loop at x of height h & width dx :

E
d r   E  x , t  h  E  x  dx , t  h


E
  E  x, t  h   E  x, t  
d x h
x


i
V 
j
k



 x
 y
 z
Vx
Vy
Vz
 B  B h dx
Faraday’s Law :
Faraday’s law expressed as a differential eq :
dB
dt
E
x

B
t
 
E  
h dx
E  E yˆ
 E

 
d x h
 x

B  B zˆ
B
Ey
t
x
B
t
 
 Bz
t
Ampere-Maxwell Law

B
C
dr  0  0
dE
I=0
dt
For loop at x of height h & width dx :

B
C
d r  B  x  h  B  x  dx  h
B  B zˆ


 B

B
 B  x, t  h   B  x, t  
d x h   
d x h
x


 x

dE
 E  E h dx
Ampere-Maxwell Law :

B
x
dt
 0  0
Ampere-Maxwell law expressed as a differential eq :
E
t

E
t
E  E yˆ
h dx

  B  0  0
 Bz
x
E
t
  0 0
Ey
t
in vacuum
Conditions on Wave Fields
For E = E(x,t) j & B = B(x,t) k,
Faraday’s Law :
Ampere-Maxwell Law :
For a plane wave
Faraday’s Law :
Ampere-Maxwell Law :
E
x

B
x
 
B
V 
t
 0  0
i
j
k



 x
 y
 z
Vx
Vy
Vz
E
t
E  x , t   E p sin  kx   t  ˆj
B  x , t   B p sin  kx   t  kˆ
k E p c os  k x   t    B p c os  k x   t 

k Ep   Bp
 k B p c os  k x   t     0  0  E p c os  k x   t 

k B p  0  0  E p
29.5. Properties of Electromagnetic Waves
k B p  0  0  E p
k Ep   Bp

k
speed of wave =

2
 0  0 

k
1
0  0
 3  10 m / s
8
2
1

 4
 10
7
N / A
2
  4

2
2 
C / N m 
 9  10

1
= speed of light in vacuum
9
=c
Maxwell: light is EM wave.
1983: meter is defined so that c is exactly 299,792,458 m/s.
Hence, 0 = 1 / (4  c2  107 ) C2/Nm2, where c = 299,792,458.
GOT IT? 29.1
At a particular point E of an EM wave points in the +y direction,
while B points in the z direction.
Is the propagation direction
(a) +x ;
(b) x ;
(c) either +x or x but you can’t tell which;
(d) y ;
(e) +z ; or
(f) not along any of the coordinate axes ?
Example 29.2. Laser Light
A laser beam with wavelength 633 nm is propagating through air in the +z direction.
Its electric field is parallel to the x axis and has magnitude 6.0 kV/m.
(a)
the wave frequency,
(b)
the amplitude of the magnetic field, and
(c)
the direction of the magnetic field.
(a)
(b)
f 
B 


c
3.0  10 m / s
8
c
E
Find
633  10
9
 4.7  10
14
m
6.0  10 V / m
3

3.0  10 m / s
8
 2.0  10
 20  T
(c)
Hz
y axis.
5
T
Polarization
Polarization  // E.
Radiation from antennas are polarized.
E.g., radio, TV, ….
εˆ  ˆj
Light from hot sources are unpolarized.
E.g., sun, light bulb, …
Reflection from surfaces polarizes.
E.g., light reflecting off car hoods is partially polarized in
horizontal direction.
εˆ  kˆ
Transmission through crystal / some plastics polarizes.
E.g., Polaroid sunglasses, …
Only component of E // preferred direction e is transmitted.
E tra n s   E in c  εˆ  εˆ
Law of Malus :
 E inc cos  εˆ
E trans
2
 E inc
2
 = angle between Einc & .
cos 
2
or
S trans  S inc cos 
2
S  S 0 cos 
2
2 polarizers with mutually perpendicular transmission axes.
No light gets through where they overlap.
Polarization of EM wave gives info about its source & the medium it passes through.
Applications: astronomy, geological survey, material stress analysis, …
Liquid crystal display (LCD)
Unpolarized
light
Horizontal polarizer
passes light.
Vertical polarizer
passes only Ev .
LC molecules rotate
polarization to horizontal
direction.
Horizontal polarizer
blocks light.
Applied V aligns molecules;
polarization not rotated.
Conceptual Example 29.1. Crossed Polarizers
Unpolarized light shines on a pair of polarizers with their transmission
axes perpendicular, so no light gets through the combination.
What happens when a third polarizer is sandwiched in between, with its
transmission axes at 45 to the others?
1st & middle polarizers not  so some light passes through.
Passed light’s polarization not  to axis of last polarizer so
some light passes through.
Making the Connection
How does the intensity of light emerging from this polarizer “sandwich”
compare with the intensity of the incident unpolarized light?
S trans  S inc cos 
2
Intensity of light emerging from 1st polarizer :
S 1  S in c
1
2
2

c os  d 
2

1
2
0
S inc
( polarized along axis of 1st polarizer )
Intensity of light emerging from middle polarizer :
S 2  S 1 cos 45 
2

1
2
S1 
1
4
S inc
( polarized along axis of middle polarizer.)
Intensity of light emerging from ensemble :
S 3  S 2 cos 45 
2

1
2
S2 
1
8
S in c
( polarized along axis of 3rd polarizer.)
29.6. The Electromagnetic Spectrum
Earth’s atmosphere:
Transparent to:
most radio, visible light.
Opaque to:
most IR, upper UV, X-rays,  rays.
UV is absorbed by ozone layer
IR by green house gases.
29.7. Producing Electromagetic Waves
Any changing E or B will create EM waves.
Any accelerated charge produces radiation.
Radio transmitter: e’s oscillate in antenna driven by LC circuit.
X-ray tube: accelerated e’s slammed into target.
MW magnetron tube: e’s circle in B.
EM wave :
f = f of q motion
Most efficient:  ~ dimension of emitter / reciever
Outgoing
EM waves
Source
replenishes
radiated
energy
LC oscillator
drives I in
antenna
Waves emit / receive  axis of dipole.
29.8. Energy & Momentum in Electromagetic Waves
Consider box of thickness dx, & face A  k of EM wave.
uE 
Energy densities:
1
0 E
2
2
uB 
1
2 0
B
2
Energy in box:
dU   uE  uB  A dx
Rate of energy moving through box:
dU
dt
Intensity S = rate of energy flow per unit area


1 
1
2
2 

E

B
 0
 A dx
2 
0

d U

dx / c
S 
1 
1
2
2 

E

B
 0
 Ac
2 
0

1 
1
2
2 

E

B
 0
c
2 
0

Plane waves:
B 
E
c

1 
1 
S 
c



 E B c
0
2 
0 c 

1
20
c
2
 00  1
E
B

1
0
E B
1
S 
Plane waves:
S 
In general:
E B
0
1
0
EB
Poynting vector
Average intensity for plane waves :
S 
1
0
E B

2

1 E pk
0 2 c

1 E pk B pk
0
2
1
B pk
0
2
c
2
E, B in phase
see Prob 64
GOT IT? 29.3
Lasers 1 & 2 emit light of the same color,
& E in the beam from laser 1 is twice as strong as that in laser 2’s beam.
How do their
B1 = 2 B 2
(a)
magnetic fields,
S1 = 4 S 2
(b)
intensities, and
1 = 2
(c)
wavelengths compare?
Example 29.3. Solar Energy
The average intensity of noontime sunlight on a clear day is about 1 kW/m2.
(a) What are the peak electric & magnetic fields in sunlight ?
(b) At this intensity, what area of 40% efficient solar collectors would you need to
replace a 4.8-kW water heater ?
(a)
S 
1
0
2
c
B pk
2
B pk 
20 S

2  4  10
6
Area needed is
 2.9  T
T
4.8 kW
40%  1.0 kW / m
3
8
E pk  c B pk  8.7  10 2 V / m
(b)
H / m   10 W / m
3  10 m / s
c
 2.9  10
7
2
 12 m
2
 0.87 kV / m
2

Waves from Localized Sources
Afar from localized source, wave is spherical :
S  E ,B
2
2

E ,B 
S 
P
4 r
2
1
r
 wave fields dominates static fields away from the sources.
Intensity = power / area
Example 29.4. Cell Phone Reception
A cell phone’s typical average power output is about 0.6 W.
If the receiver at a cell tower can handle signals with peak electric fields as weak as 1.2 mV/m,
what is the maximum allowable distance from cell phone to tower ?
S 

r 
2
P
4 r
2

1 E pk
0 2 c
2 0 c P
4  E pk
2

2  4  1 0
7
H / m   3  1 0 m / s   0 .6 W
4   1 .2  1 0
8
3
V / m
2

 5 km
Application: Cell Phones
Hexagonal cell area  25 km2.
~ circle of radius
25  10 m
6
r 

2
 2.8 km
Transmission & reception are at different frequencies.
Momentum & Radiation Pressure
Maxwell :
p ra d 
Prad 
1
U Sˆ
radiation momentum
c
S
radiation pressure
on absorbing surface
1
radiation pressure
on reflecting surface
c
Pra d  2
Cosmos 1, a solar light-sailing spacecraft,
failed at launch in 2005.
1
c
S